A) K
B) \[\frac{K}{4}\]
C) \[\frac{K}{2}\]
D) zero
Correct Answer: C
Solution :
Phase difference,\[\phi =\left( \frac{2\pi }{\lambda } \right)\]path difference For path difference \[\lambda ,\phi =\left( \frac{2\pi }{\lambda } \right)\lambda =2\pi \] For path difference \[\frac{\lambda }{4},\phi =\left( \frac{2\pi }{\lambda } \right)\frac{\lambda }{4}=\frac{\pi }{2}\] Intensity on screen \[I=4{{I}_{0}}{{\cos }^{2}}\left( \frac{\phi }{4} \right)\] For \[\phi =2\pi ,K=4{{I}_{0}}{{\cos }^{2}}\left( \frac{2\pi }{4} \right)\Rightarrow K=4{{I}_{0}}\] \[I=K{{\cos }^{2}}\frac{\phi }{2}=K{{\cos }^{2}}\left[ \frac{2\pi }{\lambda }\times \frac{\lambda }{4}\times \frac{1}{2} \right]\] \[=K{{\cos }^{2}}\frac{\pi }{4}=\frac{K}{2}\] So intensity at given point where path difference is \[\frac{\lambda }{4}\]will be \[\frac{K}{2}.\]
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